1. **Problem Statement:** Two long parallel wires carry currents $I$ and $2I$ upward. Point $P$ is midway between them, each wire at distance $d$ from $P$. The magnetic flux density at $P$ due to wire 1 is $B$. We need to find the resultant magnetic flux density magnitude and direction at $P$. 2. **Formula:** The magnetic flux density at distance $r$ from a long straight wire carrying current $I$ is given by: $$B = \frac{\mu_0 I}{2\pi r}$$ where $\mu_0$ is the permeability of free space. 3. **Given:** - Magnetic field at $P$ from wire 1 is $B$. - Current in wire 2 is $2I$. - Distance from each wire to $P$ is $d$. 4. **Calculate magnetic field at $P$ from wire 2:** Since $B_1 = \frac{\mu_0 I}{2\pi d} = B$, then $$B_2 = \frac{\mu_0 (2I)}{2\pi d} = 2B$$ 5. **Direction of magnetic fields:** Using the right-hand rule for currents upward: - Magnetic field from wire 1 at $P$ points **perpendicular into the page**. - Magnetic field from wire 2 at $P$ points **perpendicular out of the page**. 6. **Resultant magnetic field at $P$:** Since the two fields are opposite in direction, the resultant magnitude is: $$B_{\text{resultant}} = |B_2 - B_1| = |2B - B| = B$$ 7. **Direction of resultant field:** Since $B_2 > B_1$ and $B_2$ points out of the page, the resultant magnetic field at $P$ is **perpendicular out of the page**. **Final answer:** The resultant magnetic flux density at point $P$ is $B$ perpendicular out of the page.